NEET 2017 Biology Question Paper with Answer and Solution

(D) : Halophiles are a specialized group of archaebacteria that are adapted to survive in extreme saline environments,such as salt pans,salt beds,and salt marshes. They possess unique cell wall structures that allow them to thrive in high salt concentrations where other organisms cannot survive.

(C) Viroids are infectious agents that consist only of a short strand of circular,single-stranded $RNA$ without a protein coat (capsid).
In contrast,viruses consist of genetic material ($DNA$ or $RNA$) enclosed within a protein coat called a capsid.
Therefore,the defining characteristic of viroids is that they are $RNA$ molecules without a protein coat.

(B) $Mycoplasma$ are the smallest known living cells that lack a definite cell wall.
They are pathogenic to both plants and animals.
These organisms are capable of surviving without oxygen (facultative or obligate anaerobes).

(C) : The $Glycocalyx$ is the outermost layer of the bacterial cell envelope. It is composed of non-cellulosic polysaccharides and may contain proteins. This layer provides a sticky or slimy character to the bacterial cell,which helps in attachment to surfaces and protects the cell from phagocytosis.

(A) $Volvox$ is a green alga that forms spherical colonies. Each colony consists of a large number of cells embedded in a gelatinous matrix. These colonies are known as coenobia. In contrast,$Ulothrix$ is filamentous,$Spirogyra$ is also filamentous,and $Chlorella$ is a unicellular alga.

(D) $Pinus$ is a monoecious plant,$i.e.$,in $Pinus$ the male and female cones or strobili are borne on the same plant. Therefore,the option $Pinus-Dioecious$ is a mismatch.

(C) The correct answer is $C$. In $Chlamydomonas$,the life cycle is of the haplontic type. In this type of life cycle,the dominant phase is the haploid gametophyte. The zygote is the only diploid stage,which undergoes meiosis to produce haploid zoospores. Therefore,zygotic meiosis is a characteristic feature of $Chlamydomonas$.

(B) $Ectocarpus$ exhibits a haplodiplontic life cycle,where both haploid and diploid phases are multicellular and often free-living.
$Fucus$ exhibits a diplontic life cycle,where the diploid sporophyte is the dominant,photosynthetic phase,and the gametophytic phase is represented by single to few-celled gametophytes.

(B) The correct answer is $B$.
An important characteristic that hemichordates and chordates share is the presence of pharyngeal gill slits.
In hemichordates,these gill slits are dorsal in position,whereas in chordates,they are lateral.
$A$ true notochord is absent in hemichordates; instead,they possess a structure called the stomochord.
The nervous system in hemichordates is intraepidermal,and they possess a ventral nerve cord,which differs from the dorsal hollow nerve cord found in chordates.

(B) The correct answer is $B$.
Whales,dolphins,and seals are classified as aquatic mammals because they are warm-blooded,breathe air through lungs,and give birth to live young.
Trygon (stingray) and sharks belong to the class Chondrichthyes,which are cartilaginous fishes,not mammals.

(A) : $Perissodactyla$ represents the order of the horse.
$Equidae$ is the family,$caballus$ is the subspecies,whereas $E. ferus$ is the species of the horse.

(B) : The spongocoel is the central body cavity found in sponges ($Phylum$ $Porifera$). This cavity is lined by highly specialized flagellated cells known as choanocytes or collar cells. These cells play a crucial role in creating water currents and capturing food particles.

(D) The correct answer is $(d)$.
Coconut fruit is a fibrous drupe.
In a drupe,the pericarp is well-differentiated into an outer thin epicarp,a middle fleshy or fibrous mesocarp,and an inner stony hard endocarp.
In coconut,the mesocarp is fibrous,which is why it is classified as a fibrous drupe.

(B) In $Bougainvillea$,the axillary buds of the stem get modified into woody,straight,and pointed structures called thorns. These thorns provide protection to the plant against herbivores. Therefore,thorns in $Bougainvillea$ are modifications of the stem.

(B) The vascular cambium is a lateral meristem that undergoes periclinal divisions.
These divisions occur on both the inner and outer sides of the cambium ring.
The cells produced towards the inner side differentiate into secondary xylem,while the cells produced towards the outer side differentiate into secondary phloem.
Therefore,the vascular cambium gives rise to secondary xylem and secondary phloem.

(D) : Root hair are lateral tubular outgrowths that develop from the outer cells of the zone of maturation or root hair zone. This region is characterized by the presence of epidermal cells that differentiate into root hairs to increase the surface area for water and mineral absorption.

(B) $Phellem$ (cork) is composed of dead cells that are compactly arranged and rectangular in shape. These cells have suberized cell walls,which make them impermeable to water and gases.

(B) : Heartwood is the non-functional part of secondary xylem. It consists of dead elements with highly lignified walls and is filled with organic compounds like tannins,resins,oils,and gums,which make it highly durable and resistant to microbial attack. Because its vessels are blocked by these deposits,it does not conduct water and minerals.

(C) The correct route for the transport of sperms in a male frog is:
Testes $\rightarrow$ Vasa efferentia $\rightarrow$ Kidney $\rightarrow$ Bidder’s canal $\rightarrow$ Urinogenital duct $\rightarrow$ Cloaca.
In male frogs,sperms are produced in the testes. They travel through $10-12$ vasa efferentia into the kidneys. Within the kidneys,they enter the Bidder’s canal,which communicates with the urinogenital duct. Finally,the sperms are released into the cloaca and then to the exterior.

(C) The correct answer is $(C)$.
The heart of a frog is myogenic, which means the heartbeat originates from the specialized cardiac muscle cells rather than from nervous impulses.
Because the heart is myogenic, it is also autoexcitable, meaning it can generate its own electrical impulses to initiate contraction.
Due to these properties, the heart continues to beat outside the body as long as it receives an adequate supply of oxygen, nutrients, and $ATP$ to sustain the metabolic activity of the cardiac muscle cells.

(C) : Mitochondria are miniature biochemical factories where food substances or respiratory substrates are completely oxidized to carbon dioxide and water.
The energy liberated in this process is initially stored in the form of reduced coenzymes and reduced prosthetic groups.
These subsequently undergo oxidation to form energy-rich $ATP$.
$ATP$ is released from the mitochondria and helps perform various energy-requiring processes of the cell,such as muscle contraction,nerve impulse conduction,biosynthesis,membrane transport,cell division,and movement.
Because of the formation of $ATP$,mitochondria are known as the powerhouses of the cell.

(A) The correct statement is $Holoenzyme = Apoenzyme + Cofactor$.
An apoenzyme is the protein part of an enzyme,which is inactive on its own.
$A$ cofactor is a non-protein chemical compound that is required for the enzyme's activity.
When an apoenzyme binds with its specific cofactor,it forms a complete,catalytically active enzyme known as a $Holoenzyme$.
Coenzymes are a specific type of organic cofactor that bind loosely to the apoenzyme.
Therefore,$Holoenzyme = Apoenzyme + Coenzyme$ (where coenzyme acts as a cofactor) is the correct representation among the given options.

(C) The correct answer is $C$.
Lipids are generally fatty acid esters of alcohols and related substances,and they do not form long chains of repeating monomeric units in the same way as other macromolecules.
Polysaccharides are polymers of monosaccharides.
Proteins are polymers of amino acids.
Nucleic acids are polymers of nucleotides.

(A) The correct sequence of events during mitosis is:
$1$. Condensation: During prophase,chromatin condenses into compact chromosomes.
$2$. Nuclear membrane disassembly: Occurs during prometaphase.
$3$. Arrangement at equator: Chromosomes align at the metaphase plate during metaphase.
$4$. Centromere division: Occurs at the start of anaphase.
$5$. Segregation: Sister chromatids move to opposite poles.
$6$. Telophase: Chromosomes decondense and the nuclear envelope reforms.
Crossing over is a characteristic feature of meiosis,not mitosis. Therefore,option $(A)$ provides the most complete and accurate sequence of events.

(B) $APC$ (Anaphase Promoting Complex) is an E3 ubiquitin ligase that triggers the transition from metaphase to anaphase by tagging specific proteins (like securin and cyclin $B$) for degradation.
Securin degradation releases the enzyme separase,which cleaves the cohesin proteins holding sister chromatids together.
If $APC$ is defective,these proteins are not degraded,cohesin remains intact,and sister chromatids fail to separate.
Therefore,the chromosomes will not segregate to opposite poles.

(D) By convention, the water potential of pure water at standard temperatures, which is not under any pressure, is taken to be zero.
Water potential is denoted by the Greek symbol $\Psi_w$ (psi).
Since pure water has the highest concentration of water molecules, it has the highest water potential, which is defined as $0$.

(B) The correct answer is $B$.
When the turgidity of the two guard cells flanking each stomatal aperture increases,the thin outer walls bulge out,forcing the thick inner walls into a crescent shape.
This mechanical change results in the opening of the stomata.
The opening process is specifically facilitated by the radial orientation of cellulose microfibrils in the cell walls of the guard cells.
This radial arrangement prevents the cells from expanding in diameter,forcing them to expand in length and curve outward,which opens the pore.

(B) The correct answer is $B$.
Statement $B$ is incorrect because $C_4$ plants show higher rates of photosynthesis at higher temperatures,whereas $C_3$ plants have a lower temperature optimum.
$C_4$ plants are adapted to warmer environments and possess a mechanism to minimize photorespiration,allowing them to perform efficiently at higher temperatures.
Conversely,$C_3$ plants perform better at cooler temperatures.
Other statements are correct:
$1$. Increasing $CO_2$ concentration up to $0.05\%$ acts as a fertilizer for $C_3$ plants.
$2$. Tomato and bell pepper are greenhouse crops grown in $CO_2$-enriched environments.
$3$. Light saturation for $CO_2$ fixation occurs at $10\%$ of full sunlight,except for plants in shade or dense forests.

(A) In $C_4$ plants,the primary $CO_2$ acceptor is a $3$-carbon molecule called Phosphoenol pyruvate $(PEP)$.
This reaction occurs in the mesophyll cells and is catalyzed by the enzyme $PEP$ carboxylase $(PEPCase)$.
In contrast,$C_3$ plants use Ribulose $1,5$-bisphosphate $(RuBP)$ as the primary $CO_2$ acceptor.
Therefore,the correct option is $(A)$.

(C) The correct answer is $C$. The Krebs' cycle starts with the condensation of an acetyl group (acetyl $CoA$,$2C$) with oxaloacetate $(4C)$ to form a tricarboxylic,$6$-carbon compound called citric acid. It does not condense with pyruvic acid.

(B) : In low concentrations,auxins such as $2,4-D$ ($2,4$-Dichlorophenoxy acetic acid) are useful in preventing pre-harvest fruit drop and leaf drop.

(A) : Paneth cells,present in the base of the 'Crypts of Lieberkuhn',are rich in zinc and contain acidophilic granules.
These cells are known to secrete antibacterial lysozyme,which helps in maintaining the intestinal flora.
Zymogen cells (or chief cells) are found in the stomach and secrete pepsinogen.
Kupffer cells are specialized macrophages found in the liver that perform phagocytosis.
Argentaffin cells,found in the 'Crypts of Lieberkuhn',are involved in the synthesis of hormones like secretin and $5$-hydroxytryptamine.

(C) The correct option is $C$.
Pancreatic juice is secreted by the pancreas and contains several inactive enzymes (proenzymes) and active enzymes.
The key components include:
$1$. Proenzymes: $Trypsinogen$,$Chymotrypsinogen$,and $Procarboxypeptidase$.
$2$. Active enzymes: Pancreatic $\alpha$-amylase,Pancreatic lipase,$DNase$,$RNase$,and $Elastase$.
$3$. Sodium bicarbonate: To neutralize stomach acid.
Option $C$ correctly lists Lipase,Amylase,Trypsinogen,and Procarboxypeptidase,which are all essential components of pancreatic juice. Pepsin and Rennin are gastric enzymes,and Maltase is an intestinal enzyme.

(B) The dental formula for a child (deciduous or milk teeth) is $2102/2102$,which totals $20$ teeth.
In this set,there are $8$ incisors,$4$ canines,and $8$ molars.
Premolars are completely absent in the primary (milk) dentition of a two-year-old child.
Therefore,the correct answer is $B$.

(D) The correct answer is $D$.
Residual volume is the volume of air that remains in the lungs even after a forceful expiration.
This volume of air ensures that the lungs do not collapse during the process of breathing.
It also allows for the continuous exchange of gases between the blood and the alveoli even when no more air can be exhaled.

(D) The correct answer is $(d)$.
Adult human red blood cells $(RBCs)$ are enucleated,meaning they lack a nucleus and other cell organelles like mitochondria,ribosomes,and Golgi bodies.
The primary advantage of this structural adaptation is that it maximizes the internal space within the cell.
By removing the nucleus and organelles,the cell can accommodate a significantly larger amount of haemoglobin,which is the oxygen-carrying pigment.
This allows for more efficient transport of oxygen throughout the body.

(D) : Blood enters the liver from two sources. The hepatic artery provides oxygenated blood,while the hepatic portal vein carries deoxygenated blood. The blood in the hepatic artery originates from the aorta. The blood in the hepatic portal vein comes directly from the digestive tract,including the intestine and stomach,containing newly absorbed nutrients.

(A) The correct answer is $(A)$.
Atrial natriuretic factor $(ANF)$ is released by the atrial walls of the heart in response to an increase in blood volume and pressure.
It acts as a vasodilator and promotes the excretion of sodium and water,thereby lowering blood pressure and volume.
Conversely,a decrease in blood pressure or volume triggers the release of renin (from $JGA$),$ADH$ (from the posterior pituitary),and aldosterone (from the adrenal cortex) to restore homeostasis.
Therefore,a decrease in blood pressure or volume will not cause the release of $ANF$.

(D) The correct statement is that the ascending limb of the loop of Henle is impermeable to water.
$1$. The descending limb of the loop of Henle is permeable to water but impermeable to electrolytes.
$2$. The ascending limb of the loop of Henle is impermeable to water but permeable to electrolytes (solutes like $Na^+$,$Cl^-$).
Therefore,option $D$ is the correct statement.

(B) The joint between the $atlas$ (first cervical vertebra) and the $axis$ (second cervical vertebra) is known as a $pivot$ $joint$.
$Pivot$ $joints$ are a specific category of $synovial$ $joints$ that allow for rotational movement,such as the side-to-side rotation of the head.
Therefore,the correct option is $(b)$.

(D) Humans have $12$ pairs of ribs.
These are classified based on their attachment to the sternum.
The first $7$ pairs of ribs are called 'true ribs'.
They are attached dorsally to the thoracic vertebrae and ventrally connected to the sternum with the help of hyaline cartilage.
Therefore,$X = 12$ and $Y = 7$.

(D) : Schwann cells and oligodendrocytes form the myelin sheath around the axon.
Myelin sheath acts as an insulating layer,which prevents the loss of energy of the nerve impulse during its transmission along the nerve fiber.

(C) The correct answer is $C$.
Neurotransmitters are chemical substances responsible for the transmission of nerve impulses across a synapse.
These chemicals are released from the synaptic vesicles into the synaptic cleft.
Once in the cleft,the neurotransmitters bind to specific protein receptor molecules located on the post-synaptic membrane.
This binding causes the opening of ion channels,leading to the depolarization of the post-synaptic membrane and the generation of a new action potential.

(A) Statement $(1)$ is correct because carotene is a precursor to vitamin $A$.
Statement $(2)$ is incorrect because photopigments are embedded in the membrane discs of the outer segment,not the inner segment.
Statement $(3)$ is correct because retinal is an aldehyde derivative of vitamin $A$.
Statement $(4)$ is correct because retinal is the light-absorbing component of all visual photopigments (rhodopsin/iodopsin).
However,looking at the provided options,the most accurate combination regarding the specific biological role of carotene and vitamin $A$ derivatives in vision is $(1), (3)$ and $(4)$. Since the provided solution was $(b)$,but $(4)$ is scientifically correct,the best choice is $(A)$.

(A) $GnRH$ (Gonadotropin-releasing hormone) is a peptide hormone produced by the hypothalamus.
It travels through the hypophyseal portal system to reach the anterior pituitary gland.
Upon reaching the anterior pituitary,it stimulates the gonadotroph cells to synthesize and release two key gonadotropins: Luteinizing Hormone $(LH)$ and Follicle-Stimulating Hormone $(FSH)$.
These hormones are essential for the regulation of reproductive processes in both males and females.

(A) : Epiphyseal plate is a hyaline cartilage plate in the metaphysis at each end of a long bone. It is the region of the long bone where new bone growth occurs during development. In adults,elevated levels of $GH$ result in acromegaly,a condition characterized by the thickening of bones rather than an increase in height,because the epiphyseal plates have already ossified (closed) after adolescence.

(A) $Rhodospirillum$ is a free-living nitrogen-fixing bacterium.
Mycorrhiza is the symbiotic association between fungi and the roots of higher plants.
The most common fungal partners of mycorrhiza are $Glomus$ species.
Therefore,the pair $Rhodospirillum - Mycorrhiza$ is a mismatch.

(B) The plants that grow in saline,marshy,or swampy areas are called $Halophytes$.
These plants develop special roots called $pneumatophores$ (respiratory roots) that grow vertically upwards out of the soil to obtain oxygen for respiration,as the soil is waterlogged and lacks sufficient air.
Additionally,these plants exhibit $vivipary$,a phenomenon where seeds germinate while still attached to the parent plant,which is an adaptation to survive in saline environments.
Therefore,the correct answer is $Halophytes$.

(D) Mitochondria are known as the '$powerhouse$ of the cell'.
They are the sites of aerobic respiration,where they perform the oxidation of carbohydrates to release energy in the form of $ATP$ (Adenosine Triphosphate).
Lysosomes are involved in digestion,ribosomes in protein synthesis,and chloroplasts in photosynthesis.

(D) In bacteria,the glycocalyx is an outer layer composed of polysaccharides and/or polypeptides.
It provides a sticky or slimy character to the bacterial cell,which helps in adhesion to surfaces and protects the cell from phagocytosis by the host immune system.
Therefore,the correct option is $D$.

(B) The correct answer is $B$. In bacteria,$DNA$ replication occurs prior to binary fission. Bacteria are prokaryotes and do not undergo the cell cycle phases like $S$ phase,nor do they have a nucleolus. Replication is a prerequisite for the cell to divide and ensure that each daughter cell receives a complete copy of the genome.

(B) The coconut fruit is a drupe. In a coconut,the edible part is the endosperm. The coconut water represents the free-nuclear endosperm,while the white kernel represents the cellular endosperm. Therefore,the correct option is $B$.

(B) : The $Corpus$ luteum is a temporary endocrine gland found in the human female reproductive system.
It is formed from the ruptured Graafian follicle after ovulation.
It secretes the hormone progesterone, which is essential for the maintenance of the endometrium during pregnancy.
If fertilization does not occur, the $Corpus$ luteum degenerates, leading to a drop in progesterone levels and the onset of menstruation.

(B) : In angiosperms,the functional megaspore is the first cell of the female gametophyte. It enlarges and undergoes mitotic divisions to form the embryo sac,which is the mature female gametophyte.

(A) : Entomophily is the most common type of zoophily where pollination takes place through the agency of insects.
Entomophilous flowers are brightly coloured and secrete nectar to attract visiting insects.
Anemophily (wind pollination) and hydrophily (water pollination) do not require attractants or rewards due to the involvement of abiotic pollinating agents.
Cleistogamy is self-pollination in closed flowers.

(B) The correct answer is $B$. Flowers that possess a single ovule in each ovary and are arranged in a compact inflorescence are typical adaptations for wind pollination (anemophily). This arrangement helps in the efficient dispersal of pollen grains by wind currents.

(A) In dioecious plants,male and female flowers are present on different plants.
Since the male and female reproductive structures are physically separated on distinct individuals,self-pollination cannot occur.
Autogamy (transfer of pollen grains from the anther to the stigma of the same flower) is prevented because the plant is unisexual.
Geitonogamy (transfer of pollen grains from the anther of one flower to the stigma of another flower on the same plant) is also prevented because there is only one sex per plant.
Therefore,dioecious plants prevent both autogamy and geitonogamy.

(C) The correct answer is $C$. Double fertilisation is a unique and characteristic feature of angiosperms.
In this process,two male gametes are released into the embryo sac.
One male gamete fuses with the egg cell to form a diploid zygote,which develops into the embryo.
The second male gamete fuses with the secondary nucleus (or two polar nuclei) to form the primary endosperm nucleus $(PEN)$,which is triploid $(3n)$ and develops into the endosperm.

(C) The correct answer is $C$. Capacitation is the physiological maturation process that spermatozoa undergo within the $female$ $reproductive$ $tract$.
During this process, secretions from the female genital tract remove coating substances (such as cholesterol and glycoproteins) deposited on the surface of the sperm, particularly over the acrosome.
This removal exposes the receptor sites on the acrosome, which allows the sperm to bind to the zona pellucida of the egg and become active enough to penetrate it.
Therefore, capacitation is essential for fertilization and occurs only after the sperm enters the female reproductive system.

(D) The correct matches are as follows:
$A$. Gonorrhoea is caused by the bacterium $Neisseria \text{ gonorrhoeae}$ $(ii)$.
$B$. Syphilis is caused by the bacterium $Treponema \text{ pallidum}$ $(iii)$.
$C$. Genital warts are caused by the $Human \text{ papilloma virus}$ $(iv)$.
$D$. $AIDS$ is caused by the $Human \text{ Immunodeficiency Virus}$ $(HIV)$ $(i)$.
Therefore, the correct matching sequence is $A-ii, B-iii, C-iv, D-i$.

(D) : Copper-releasing $IUDs$ (e.g.,$CuT$,$Cu7$,$Multiload$ $375$) are placed in the uterus of females.
These devices are highly efficient methods of birth control.
The copper ions $(Cu^{2+})$ released by these devices increase phagocytosis of sperms within the uterus.
Furthermore,these ions suppress the motility of sperms and reduce their fertilising capacity,thereby preventing fertilization.

(D) In cases where the male partner is unable to inseminate the female due to a very low sperm count,the technique of Artificial Insemination $(AI)$ is used.
In this technique,the semen collected either from the husband or a healthy donor is artificially introduced into the vagina or into the uterus (Intrauterine Insemination - $IUI$) of the female.

(B) $Sickle$ $cell$ $anaemia$ is caused due to a point mutation in which glutamic acid is replaced by valine at the $6th$ position of the $\beta$-globin chain. Thus,it is a qualitative defect in the functioning of globin molecules.
$Thalassemia$ is caused due to either mutation or deletion,which ultimately results in a reduced rate of synthesis of one of the globin chains that make up haemoglobin. Hence,it is a quantitative defect in the functioning of globin molecules.

(B) The cross between the parents with genotypes $I^AI^B$ and $I^Ai$ results in the following offspring genotypes:

$I^A I^A$	Genotype $I^A I^A$ (Phenotype $A$)
$I^A i$	Genotype $I^A i$ (Phenotype $A$)
$I^A I^B$	Genotype $I^A I^B$ (Phenotype $AB$)
$I^B i$	Genotype $I^B i$ (Phenotype $B$)

As shown above,there are $4$ distinct genotypes $(I^AI^A, I^Ai, I^AI^B, I^Bi)$ and $3$ distinct phenotypes $(A, AB, B)$.

(D) : Down's syndrome is an autosomal aneuploidy caused by the presence of an extra chromosome number $21$. During oogenesis,non-disjunction occurs,leading to both chromosomes of the $21$st pair passing into a single egg. This results in a trisomy of chromosome $21$ in the zygote.

(A) The correct answer is $A$. Mendel studied $7$ pairs of contrasting traits in pea plants. The character 'Trichomes-Glandular or non-glandular' was not one of them.

$1$. Seed shape	Round $(R)$ / Wrinkled $(r)$
$2$. Seed colour	Yellow $(Y)$ / Green $(y)$
$3$. Flower colour	Violet $(V)$ / White $(v)$
$4$. Pod shape	Inflated $(I)$ / Constricted $(i)$
$5$. Pod colour	Green $(G)$ / Yellow $(g)$
$6$. Flower position	Axial $(A)$ / Terminal $(a)$
$7$. Stem height	Tall $(T)$ / Dwarf $(t)$

(D) Gregor Johann Mendel conducted his hybridization experiments on garden pea ($Pisum$ $sativum$) for $7$ years.
These experiments were carried out during the period from $1856$ to $1863$.
Therefore,the correct option is $(d)$.

(A) The final proof that $DNA$ is the genetic material was provided by Alfred Hershey and Martha Chase in $1952$.
They worked with bacteriophages,which are viruses that infect bacteria.
They grew some bacteriophages in a medium containing radioactive phosphorus $(^{32}P)$ to label the $DNA$ and others in a medium containing radioactive sulfur $(^{35}S)$ to label the protein coat.
After allowing the phages to infect $E. coli$ bacteria,they agitated the mixture in a blender to remove the viral coats from the bacterial cells.
They found that radioactive phosphorus $(^{32}P)$ was present inside the bacterial cells,indicating that $DNA$ had entered the cells,while radioactive sulfur $(^{35}S)$ remained outside,indicating that proteins did not enter the cells.
This confirmed that $DNA$ is the genetic material.

(B) The genetic code is a triplet,meaning $1$ codon consists of $3$ bases.
When a base at position $901$ is deleted,it causes a frameshift mutation from that point onwards.
The total number of bases remaining after the deletion is $998$.
The number of bases from the point of deletion $(901)$ to the end of the $RNA$ is $999 - 901 + 1 = 99$ bases.
Since each codon consists of $3$ bases,the number of codons affected by this frameshift is $99 / 3 = 33$ codons.
Therefore,$33$ codons will be altered.

(C) The correct answer is $C$.
$DNA$ replication occurs in the $5' \rightarrow 3'$ direction.
Since the two strands of the $DNA$ double helix are antiparallel,one strand (the leading strand) is synthesized continuously towards the replication fork.
The other strand (the lagging strand) is synthesized discontinuously in short segments known as Okazaki fragments.
Because the lagging strand template runs in the $5' \rightarrow 3'$ direction relative to the fork,its synthesis must proceed away from the replication fork to maintain the $5' \rightarrow 3'$ polymerization requirement.

(D) $rRNA$ (ribosomal $RNA$) is the most abundant of all types of $RNA$ $(70-80\%)$.
Hence,it is present in the highest amount in an animal cell.
The percentage of $tRNA$ is approximately $15\%$,and $mRNA$ is approximately $2-5\%$.
$miRNA$ (micro $RNA$) are short $21-22 \ bp$ long $RNA$ molecules that regulate gene expression by causing the degradation of specific $mRNA$ molecules.

(C) : Spliceosomes are complexes of small nuclear ribonucleoproteins $(snRNPs)$ that facilitate the removal of introns from pre-$mRNA$ during $RNA$ splicing.
Since bacteria are prokaryotes,their genes do not contain introns.
Consequently,prokaryotic $mRNA$ does not undergo splicing,and therefore,spliceosomes are absent in bacterial cells.

(B) $H_1$ histone is associated with the linker $DNA$ that connects adjacent nucleosomes. This binding facilitates the folding and compaction of the nucleosome chain into a higher-order structure known as the $30 \ nm$ chromatin fibre. Therefore,the presence of $H_1$ indicates that the $DNA$ is being condensed into a chromatin fibre.

(A) The correct answer is $A$.
Transplantation of tissues or organs often fails due to non-acceptance by the patient's body.
Therefore,tissue matching and blood group matching are essential before undertaking any graft or transplant.
When the immune system recognizes the proteins in the transplanted tissue or organ as foreign,it initiates a cell-mediated immune response.
This response is primarily carried out by $T$-lymphocytes,which attack the transplanted tissue,leading to the rejection of the organ.
To suppress this immune response during transplantation,immunosuppressants are administered to the patient.

(D) $MALT$ (Mucosa-Associated Lymphoid Tissue) represents significant aggregations of lymphoid tissues located in the lining of major tracts,such as the respiratory,digestive (alimentary canal),and urogenital tracts.
It constitutes approximately $50$ percent of the total lymphoid tissue in the human body.

(A) Artificial selection for a specific trait,such as higher milk yield in cows,shifts the mean value of that trait in the population towards the desired extreme.
This type of selection,where the population mean moves in one direction,is known as directional selection.
In this process,individuals with the desired trait (higher milk yield) are selectively bred,causing the frequency of alleles associated with that trait to increase over generations.
Therefore,the correct answer is $A$.

(D) The correct answer is $D$.
Inbreeding refers to the mating of more closely related individuals within the same breed for $4-6$ generations.
This process increases homozygosity,which is essential for the evolution of a pureline in any animal.
By mating related individuals,harmful recessive genes are exposed and eliminated,and desirable traits are concentrated,leading to the development of homozygous purelines.

(B) : Primary or physical treatment is the process of removal of small and large,floating and suspended solids from sewage through two processes of filtration and sedimentation.

(C) $Saccharomyces$ $cerevisiae$ (Brewer's yeast) is used for the commercial production of ethanol through fermentation.
$Methanobacterium$ is an anaerobic bacterium used in the production of biogas (methane).
$Penicillium$ $notatum$ is the fungus from which the first antibiotic,penicillin,was discovered.
$Acetobacter$ $aceti$ is a bacterium used for the production of acetic acid (vinegar).

(A) The logistic growth curve is represented by the equation $\frac{dN}{dt} = rN \left( \frac{K-N}{K} \right)$.
When the population density $(N)$ reaches the carrying capacity $(K)$,the term $\left( \frac{K-N}{K} \right)$ becomes zero.
Consequently,the growth rate $\frac{dN}{dt}$ becomes zero,and the population size remains constant.
This results in the formation of an asymptote in the growth curve,which occurs when $N = K$.

(A) : Presence of pneumatophores,$i.e.$,small negatively geotropic vertical roots,and vivipary,or seed germination while attached to the parent plant,are characteristic adaptations of halophytes. These are plants that grow in saline habitats,such as mangroves.

(A) The vertical distribution of different species occupying different levels in an ecosystem is called stratification. In a tropical rainforest,the vegetation is highly complex and shows distinct vertical layers (such as the emergent layer,canopy,understory,and forest floor) due to intense competition for sunlight and high biodiversity. Therefore,the presence of plants arranged into well-defined vertical layers is best seen in a tropical rainforest.

(C) Mycorrhizae represent a mutualistic symbiotic association between fungi and the roots of higher plants. In this relationship,the fungus helps the plant in the absorption of essential nutrients like phosphorus from the soil,while the plant provides the fungus with carbohydrates and a habitat. Since both organisms benefit from this interaction,it is classified as mutualism.

(D) The correct answer is $D$. In a forest ecosystem,the biomass is primarily contributed by large trees,which have significant woody tissue and long lifespans. This results in a much higher standing crop of biomass compared to grassland,pond,or lake ecosystems,where the primary producers are smaller plants or phytoplankton with much faster turnover rates.

(D) $Ex situ$ conservation refers to the protection and preservation of rare or threatened species of plants and animals outside their natural habitats.
This method involves maintaining these species in controlled environments such as botanical gardens, zoological parks, wildlife safari parks, and gene banks.
In contrast, biodiversity hotspots, the Amazon rainforest, and the Himalayan region are examples of $in situ$ conservation, where species are protected within their natural ecosystems.

(B) $Alexander \text{ von Humboldt}$ described the species-area relationship for the first time.
He observed that within a region, species richness increases with increasing explored area, but only up to a limit.

(D) : The $Core$ $zone$ or $Natural$ $zone$ of a biosphere reserve is an undisturbed and legally protected ecosystem.
No human activity is allowed in this zone.
Little human activity is allowed in the $buffer$ $zone$,whereas in the $transition$ $zone$,active cooperation exists between reserve management and local people for activities such as settlements and cropping.
The $restoration$ $zone$ is a degraded area selected for restoration to its near-natural form.

(B) Aerosols are fine solid or liquid particles suspended in the atmosphere. They act as environmental pollutants that negatively impact agricultural productivity by reducing sunlight reaching crops and altering precipitation patterns. Therefore,the statement that they cause increased agricultural productivity is incorrect.

(C) The $Corpus$ $luteum$ is a temporary endocrine structure in the human female reproductive system.
It is formed from the ruptured follicle after ovulation.
It secretes large amounts of progesterone, which is essential for maintaining the endometrium during pregnancy.
If fertilization does not occur, the $Corpus$ $luteum$ degenerates into the $Corpus$ $albicans$.

(D) The process of separating $DNA$ fragments based on their size is known as gel electrophoresis.
After the $DNA$ fragments are separated on an agarose gel,they cannot be seen directly under visible light.
To visualize the $DNA$ fragments,the gel is stained with a fluorescent dye called Ethidium bromide $(EtBr)$.
When the stained gel is exposed to $UV$ radiation,the $DNA$ fragments appear as bright orange-colored bands.

(A) $Selectable marker$ is a gene introduced into a cell, especially a bacterium or to cells in culture, that confers a trait suitable for artificial selection.
In recombinant $DNA$ technology, it helps in identifying and eliminating non-transformants and selectively permitting the growth of the transformants.
Examples include genes encoding resistance to antibiotics such as $ampicillin$, $chloramphenicol$, $tetracycline$, or $kanamycin$.

(B) Gel electrophoresis is a technique used to separate $DNA$ fragments based on their size.
$DNA$ molecules are negatively charged due to the phosphate groups in their backbone.
When an electric field is applied,the $DNA$ fragments move towards the anode (positive electrode).
The agarose gel acts as a molecular sieve.
Smaller $DNA$ fragments can navigate through the pores of the gel matrix more easily and quickly than larger fragments.
Therefore,smaller fragments travel a greater distance from the well compared to larger fragments.

(B) In recombinant $DNA$ technology,the production of a desired protein involves two main stages:
$1$. Upstream processing: This involves the preparation of the desired gene,selection of a suitable vector,and the cultivation of the host cells in a bioreactor to produce the protein.
$2$. Downstream processing: This stage involves the separation and purification of the expressed protein from the culture medium before it is formulated for marketing.
Therefore,the process of separation and purification is known as downstream processing.

(C) Alexander von Humboldt, a German naturalist and explorer, observed that within a region, species richness increased with increasing explored area, but only up to a limit. This relationship is described as the $Species-Area$ relationship. On a logarithmic scale, this relationship is a straight line described by the equation: $log S = log C + Z log A$, where $S$ is species richness, $A$ is area, $Z$ is the slope of the line (regression coefficient), and $C$ is the $Y$-intercept.